It suffices to show that if $A,B,C$ are three consecutive vertices of the square and $\ell$ is a line, with $X_A,X_B,X_C$ being the projections of $A,B,C$ on $\ell$, then the sum of the areas of the circles with radii $AX_A,BX_B,CX_C$ is greater than the area of $ABCD$. First of all, in order for the sum of the areas of the circles to be $<S[ABCD]$, we need all the radii $AX_A,BX_B,CX_C$ to be smaller than $\frac{\sqrt 2}2$ (assume the side of the square has length $1$). Now fix $BX_B$, and rotate $\ell$. It's easy to see that in order for the sum of the areas of the three circles to be minimal, we need $\ell$ to intersect the segments $BA,BC$, so let's assume it does, and let $T_A,T_C$ be the points of intersection. Also, let $a=BT_A,c=BT_C$. It's clear that $\sqrt{\frac 1{a^2}+\frac 1{c^2}}$ is constant, and we want to minimize $\left(\frac{AT_A}{T_AB}\right)^2+\left(\frac{CT_C}{T_CB}\right)^2=\left(\frac 1a-1\right)^2+\left(\frac 1c-1\right)^2$. This happens when $a=c$, i.e. when $\ell$ makes angles of $45^{\circ}$ with $CA,CB$, and the desired inequality is easy to get now.